## Calculus: Early Transcendentals 8th Edition

$\frac{16}{3}-4\sqrt{2}$
We begin with the integral in rectangular coordinates: $$\iint_{R}(2x-y)\,dA$$ The region in the first quadrant bounded by the circle $x^2+y^2=4$, the lines $x=0$ and $y=x$ have the bounds $0\leq r \leq2$ and $\pi/4\leq\theta\leq\pi/2$ so we can make the substitution into polar coordinates and add the bounds, getting: $$\int_{\pi/4}^{\pi/2}\bigg(\int_{0}^{2}(2r\cos\theta-r\sin\theta)\,r\,dr\bigg)d\theta$$ Cleaning this up and solving, we get: $$\int_{\pi/4}^{\pi/2}\bigg(\int_{0}^{2}r^2(2\cos\theta-\sin\theta)\,dr\bigg)d\theta\\=\int_{\pi/4}^{\pi/2}\bigg[\frac{r^3}{3}(2 \cos\theta-\sin\theta)\bigg]_{0}^{2}d\theta\\=\frac{8}{3}\int_{\pi/4}^{\pi/2}(2\cos\theta-\sin\theta)\,d\theta\\=\frac{8}{3}\bigg[2\sin\theta+\cos\theta\bigg]_{\pi/4}^{\pi/2}\\=\frac{16}{3}-4\sqrt{2}$$